physica Posted October 22, 2014 Posted October 22, 2014 I have been reading about how dipoles produce an inverse cube law. The sad thing is that I can't find any maths behind it. Am I reading a crackpot theory? If not could someone post the maths or give a link to the maths?
Carrock Posted October 22, 2014 Posted October 22, 2014 I have been reading about how dipoles produce an inverse cube law. The sad thing is that I can't find any maths behind it. Am I reading a crackpot theory? If not could someone post the maths or give a link to the maths? There is quite a lot of useful information here: https://www.physicsforums.com/threads/magnetic-force-inverse-cubed-law.587204/
swansont Posted October 22, 2014 Posted October 22, 2014 Write down the field of two monopoles of opposite charge, separated by an infinitesimal amount. Expand the denominator
Enthalpy Posted October 22, 2014 Posted October 22, 2014 The cube is correct for a dipole, I confirm. Remember this is for a static field or slowly varying. If the charges, currents... vary significantly within one propagation time between the poles, then you obtain a part as a propagating field which decreases less quickly.
physica Posted November 6, 2014 Author Posted November 6, 2014 Coming back to this topic, why would Gauss's law still be valid if Coulomb's was replaced by an inverse cube law?
Enthalpy Posted November 6, 2014 Posted November 6, 2014 It's still valid for each charge of the dipole. Only the net effect, as the difference between two forces in 1/R2, changes as 1/R3.
swansont Posted November 6, 2014 Posted November 6, 2014 Coming back to this topic, why would Gauss's law still be valid if Coulomb's was replaced by an inverse cube law? It wouldn't be. Consider a spherically symmetric scenario — the law works because the flux drops off as r^2 and so does the surface area of a sphere. That's not the case for an inverse-cube relation. IOW, E.dA is not a constant — it varies with r. 1
Enthalpy Posted November 7, 2014 Posted November 7, 2014 If the dipole consists of two finite charges with finite separation, each feels a force as 1/R2 from a distant one. In that sense, Coulomb's law still applies.
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